Question

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer 1

Hint: We will be using the concept of permutation and combination to find the total possible outcomes and total favourable outcomes then we will use the concept of probability to find the answer.
Complete step-by-step answer:
Now, we have a deck of 52 cards and we have to find the probability that both the cards are black.
Now, we know that out of 52 cards 26 cards are black and rest 26 cards are red.
Now, we know that ways of selecting a objects out of n is ${}^{n}{{C}_{a}}$ ways of selecting two black cards is ${}^{26}{{C}_{2}}$.
Total ways of selecting two cards in 52 playing cards in ${}^{52}{{C}_{2}}$.
Now, we know that,
$\begin{aligned} & P\left( Events \right)=\dfrac{\text{number of favourable outcomes}}{\text{total possible outcomes}} \\\ & P\left( 2\ black\ card \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}} \\\ \end{aligned}$
Now, we know that the value of ${}^{n}{{C}_{r}}=\dfrac{n!}{(r!)\times \left( n-r \right)!}$
$\begin{aligned} & =\dfrac{26\times 25}{52\times 51} \\\ & =\dfrac{25}{102} \\\ \end{aligned}$
Probability that both cards are black is $\dfrac{25}{102}$.

Note: To solve these types of questions it is important to note that we have used the concept of permutation and combination to find the total possible outcome and the number of favourable outcomes by using the concept that the number of ways of selecting two cards out of 26 black cards is ${}^{26}{{C}_{2}}=\dfrac{26!}{2!\times \left( 24 \right)!}$ similarly we have found the ways of selecting 2 cards out of 52 cards.